A261040 Values of c such that the Diophantine equation 5*a + 3*b = c has no solutions in positive integers.

1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 15

Offset

1,2

Comments

The minimum value of 5*a + 3*b is clearly 8 (a = b = 1), thus it is trivial that the Diophantine equation cannot be solved for any c < 8.

This sequence is finite. Proof: Given c > 7, we have three cases only to check (plus the particular cases c = 16 and c = 17 that we will see at the end), because 15 = 3 * 5. Thus, for every k > 5: 1) c := 3k --> so c = 5 * 3 + 5 * (k - 5), thus we can simply take (a = 3, b = k - 5); 2) c := 3k + 1 --> so c = 5 * 2 + 3 * (k - 3), thus (a = 2, b = k - 3); 3) c := 3k + 2 --> so c = 5 + 3 * (k - 1), thus (a = 1, b = k - 1). Now we can see that k > 5 is a sufficient but not necessary condition for the 2) and the 3), since in these cases it is sufficient that k > 3 which lead to a minimum value of 13 < 15. So we have proved that there is a solution for every c > 15.

Also, nonnegative numbers n such that the coefficient of x^n in the expansion of 1/((1-x^3)*(1-x^5))-1/(1-x^3)-1/(1-x^5) is zero. - Joerg Arndt, Aug 13 2015

Links

Table of n, a(n) for n=1..11.

E. W. Weisstein, Diophantine Equation

Example

For n = 1, 2, ..., 7 we have that 5*a + 3*b >= 8 > a(n).

Crossrefs

Sequence in context: A011870 A361202 A284831 * A087950 A060527 A152493

Adjacent sequences: A261037 A261038 A261039 * A261041 A261042 A261043

Keyword

nonn,easy,fini,full

Author

Marco Ripà, Aug 07 2015

Status

approved